|By Howdydoody (Howdydoody) on Thursday, October 09, 2003 - 08:50 pm: Edit|
This question comes from january 1999 administration, section 3, number 24.
If the 5 cards shown above are placed in a row so that is never at either end, how many different arrangements are possible.
I know the correct answer but I am having trouble understanding the logic of how many possibilities may occur at each placesetting, especially position 3 and 4. Thank you.
|By Akaflex (Akaflex) on Thursday, October 09, 2003 - 09:04 pm: Edit|
grr XIGGIIII i forgot how to solve this? u solved this b4 i forgot again
hmmm 5* 4* 3 * 2 *1 = 120 thats if for all combos..... but then u need to exclude 1 so 4! *2? then subtract? but why i wanna know WHY!!!?!?!?
|By Miseryxsignals (Miseryxsignals) on Thursday, October 09, 2003 - 09:10 pm: Edit|
120 total possibilities if any card can go anywhere. but one card (call it card A) cant be on either end. if card A is on the front then there will be 4! (4x3x2x1) different ways to arrange the other 4, same if it is on the back. so that means there are 2x4! = 48 ways to arrange the cards with A on an end. 120 - 48 = 72 possibilities. thats what i think anyways...
|By Jm405 (Jm405) on Thursday, October 09, 2003 - 09:16 pm: Edit|
fundamental counting principle
first card: 4 possibilities cuz 1 card can't be at either ends
last card: 3 possibilities
second card: 3 possibilities
3rd: 2 possibilities
fourth: 1 possibilities
4*3*3*2*1 = 72
in general, whenever restrictions are placed, always find the possibilities of them first and this is true b/c if you said 4 options for 1st card, 4 for 2nd and continued on from there, that would be wrong, cuz you could happen to lay the cards in such a fashion that the "special" card could be in the last slot, which is restricted.
hope this helps
|By Techieguy (Techieguy) on Thursday, October 09, 2003 - 09:43 pm: Edit|
I don't get the reasoning JM405.
OK I can understand why first card is 4 possibilities cuz 1 card can't be on either side. But for last card shouldn't it also be 4 cuz of the same reasoning?
|By Xiggi (Xiggi) on Thursday, October 09, 2003 - 09:58 pm: Edit|
But for last card shouldn't it also be 4 cuz of the same reasoning?
When you put the cards down, and get to the last space possible, how many cards do you have in your hand? Only one.
To illustrate, the 4*3*3*2*1 = 72, let's look at the LAST two numbers 2*1. The reason why you have 2 for the penultimate 94th card) is that you have 2 possible cards and 2 spots. You could go D - E or E - D, or two choices. As soon as you drop the 4th card, you will only hold one card and have one spot open. Hence the 1 choice.
|By Xiggi (Xiggi) on Thursday, October 09, 2003 - 10:05 pm: Edit|
Another way to look at the problem is to start with the unrestricted cards. You have 4 cards that could go anywhere.
The number of possibilities are 4! or 4 x 3 x 2 x 1 and that is 24. Easy enough but we have 5 cards and the fifth card could go in three positions.
If we imagine that the 4 cards are on the table, we know that there are 24 ways to arrange them. If we would INSERT the fifth card in an INSIDE spot, it would not increase the number of possibilities of the 4 cards but maintain the same number. However, since there are 3 spots for the fifth card, we need to consider that it would make 3 different arrangements of five cards. Hence the total number is 3 times 24 or 72.
|By Doofus (Doofus) on Thursday, October 09, 2003 - 10:05 pm: Edit|
|By Techieguy (Techieguy) on Thursday, October 09, 2003 - 10:14 pm: Edit|
I have another SAT question that includes probability. I don't get probability at all.
9. Carol wants to arrange 3 of her 4 plants in a row on a shelf. If each of the plants is in a different-colored container, how many different arrangements can she make?
And I also have another SAT problem but is unrelated to probability. If its off-topic u don't have to answer it.
25. One side of a triangle has length 6 and a second side has length 7. Which of the following could be area of this triangle?
|By Akaflex (Akaflex) on Thursday, October 09, 2003 - 10:25 pm: Edit|
24? 6Ways u can arrange 3 plants I named them A B C D ABC, BAC, CBA, ACB, CAB, BCA. But now u need to include D so there will be 6 * 4 ways? is that correct?
|By Techieguy (Techieguy) on Thursday, October 09, 2003 - 10:32 pm: Edit|
The answer is C, but is ur reasoning correct?
|By Doofus (Doofus) on Thursday, October 09, 2003 - 10:39 pm: Edit|
Yea techie 4*3*2... In the situation you just dont multiply by one which doesnt matter anyways. But if it was say 3 of 5 plants it would be 5*4*3. On #25, i dont know But II is definately a possibility. I really dont understand those problems with triangle area and sides
|By Techieguy (Techieguy) on Thursday, October 09, 2003 - 10:47 pm: Edit|
Lemme get this straight... If you had 4 of 6 plants you would do 6 x 5 x 4 x 3. If you had 2 of 3 plants you would do 3 x 2. Am I correct?
|By Xiggi (Xiggi) on Thursday, October 09, 2003 - 11:58 pm: Edit|
For the triangle problem, you know that you could have a right triangle with base = 7 and height being 6. The area would be 7*6/2 or 21.
Obviously, you could build a triangle that has a smaller area, so 13 is possible.
But, how about 24, is that possible? If 21 was a RIGHT triangle, could you build a larger triangle using sides of 6 and 7? The answer is NO but I'll let you look for the reason.
|By Metopiccockflop (Metopiccockflop) on Friday, October 10, 2003 - 12:11 am: Edit|
Triangle Inequality Theorem states the length of one side of a triangle must be greater than the difference and less than the sum of the lengths of the other two sides. Using this one can conclude that the third side a triangle with other sides measuring 6 and 7 would have to be greater than 1(7-6=1) and less than 13(7+6=13).
|By F77 (F77) on Friday, October 10, 2003 - 02:50 pm: Edit|
The triangle answer is 21. The formula for solving the area of a triangle is A=1/2B*H
3.5*6 or 3*7 both equal 21.
|By Xiggi (Xiggi) on Friday, October 10, 2003 - 03:26 pm: Edit|
Why not 13?
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